Thermodynamics - A Real Pressure Cooker

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SUMMARY

This discussion focuses on solving two thermodynamics problems involving ideal gas behavior. The first problem calculates the net force on a pressure cooker lid when heated air inside reaches 120ºC, requiring the application of the ideal gas law and consideration of atmospheric pressure. The second problem involves determining the final volume of nitrogen gas after an isobaric expansion, starting from an initial pressure of 3.0 atm and a temperature of 20ºC. Key corrections include ensuring consistent units and proper application of the ideal gas law, particularly in calculating initial and final volumes.

PREREQUISITES
  • Understanding of the ideal gas law (pV = nRT)
  • Knowledge of thermodynamic principles, specifically isobaric processes
  • Familiarity with unit conversions, particularly temperature (Celsius to Kelvin)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Review the ideal gas law applications in thermodynamics
  • Learn about isobaric processes and their implications in gas behavior
  • Study unit conversion techniques, especially for temperature and pressure
  • Explore common mistakes in thermodynamic calculations and how to avoid them
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics, as well as educators seeking to clarify common misconceptions in gas law applications.

UltraC
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Ok I am getting frustrated by these two questions in my physics assignment that seem simple but I just can't get the correct answer for.

1. If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force on the lid when the air inside the cooker had been heated to 120ºC? Assume that the temperature of the air outside the pressure cooker is 20ºC (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be P_a.
Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=NRT.


So I found an equation in my textbook F=pA
I am looking for force and I know the area so then I just have to calculate the pressure.
since v, N, and R constant:
(Pi/Ti) = (Pf/Tf)
Pf = (Pi*Tf)/(Ti) or Pf = Pi*(Tf)/(Ti) and if I know the that Ti = 20ºC(293K) and Tf 120ºC(393K)
then Pf = Pi*(393K/293K) where pi = P_a according to the question then my answer should be:
F=-(P_a*(393K/293K))*A
F=-(P_a*1.34)*A
However when I enter that as my answer it says that I am off by a multiplication factor.

2. Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20ºC undergo an isobaric expansion until the volume has tripled. What is the gas volume after the expansion?

So I started off by calculating N (number of mols)
5.0g *(1mol/28.0g) = 0.179mol

Then I used Pv=NRT to solve for the initial volume
v=1.43*10^-3cubic meters

Then P, N, and R constant:
(vi/Ti) = (vf/Tf)
vf=(vi*Tf)/Ti
vf=((1.43*10^-3)*879)/293
vf=4.29*10^-3cubic meters

However when I enter that as my answer it says that I need to check my signs.

Any help to as where I might have gone wrong would be highly appreciative.
 
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UltraC said:
Ok I am getting frustrated by these two questions in my physics assignment that seem simple but I just can't get the correct answer for.

1. If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force on the lid when the air inside the cooker had been heated to 120ºC? Assume that the temperature of the air outside the pressure cooker is 20ºC (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be P_a.

F=-(P_a*1.34)*A
However when I enter that as my answer it says that I am off by a multiplication factor.
You have to take into account the air pressure in the room that is pressing down on the lid. The net force is the F you found less P_aA. So F_net = .34 P_aA

2. Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20ºC undergo an isobaric expansion until the volume has tripled. What is the gas volume after the expansion?

So I started off by calculating N (number of mols)
5.0g *(1mol/28.0g) = 0.179mol

Then I used Pv=NRT to solve for the initial volume
v=1.43*10^-3cubic meters

Then P, N, and R constant:
(vi/Ti) = (vf/Tf)
vf=(vi*Tf)/Ti
vf=((1.43*10^-3)*879)/293
vf=4.29*10^-3cubic meters
You don't really need to involve temperature here, although it is correct. You just need to multiply vi by 3. You have done it correctly but there may be a slight rounding error. I get 4.30e-3 using 1 atm = 101325 Pa.

AM
 



It seems like you have the right approach for both questions, but there may be some small errors in your calculations. Here are some tips to help you check your work:

1. For the first question, make sure you are using the correct units for temperature. The ideal gas law uses Kelvin, not Celsius. So your calculation for Pf should be Pf = Pi*(393K/293K). Also, make sure your units are consistent throughout the equation. For example, if you are using atmospheric pressure (P_a), then your final answer should be in units of force per unit area (i.e. N/m^2 or Pa).

2. For the second question, double check your calculation for initial volume (vi). It should be vi = NRT/Pi, not vice versa. Also, make sure you are using the correct value for R. The value for R depends on the units you are using for pressure and volume. If you are using atm and cubic meters, then R = 0.0821 L*atm/mol*K. If you are using Pa and cubic meters, then R = 8.314 J/mol*K.

Overall, it seems like you have a good understanding of the concepts and equations involved in these problems. Just be careful with your units and double check your calculations to ensure accuracy.
 

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